1589. Maximum Sum Obtained of Any Permutation

Description

We have an array of integers, nums, and an array of requests where requests[i] = [starti, endi]. The ith request asks for the sum of nums[starti] + nums[starti + 1] + ... + nums[endi - 1] + nums[endi]. Both starti and endi are 0-indexed.

Return the maximum total sum of all requests among all permutations of nums.

Since the answer may be too large, return it modulo 109 + 7.

 

Example 1:

Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
Output: 19
Explanation: One permutation of nums is [2,1,3,4,5] with the following result: 
requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8
requests[1] -> nums[0] + nums[1] = 2 + 1 = 3
Total sum: 8 + 3 = 11.
A permutation with a higher total sum is [3,5,4,2,1] with the following result:
requests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11
requests[1] -> nums[0] + nums[1] = 3 + 5  = 8
Total sum: 11 + 8 = 19, which is the best that you can do.

Example 2:

Input: nums = [1,2,3,4,5,6], requests = [[0,1]]
Output: 11
Explanation: A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11].

Example 3:

Input: nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]]
Output: 47
Explanation: A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10].

 

Constraints:

• n == nums.length
• 1 <= n <= 105
• 0 <= nums[i] <= 105
• 1 <= requests.length <= 105
• requests[i].length == 2
• 0 <= starti <= endi < n

Solution

maximum-sum-obtained-of-any-permutation.py

class Solution:
    def maxSumRangeQuery(self, nums: List[int], requests: List[List[int]]) -> int:
        M = 10 ** 9 + 7
        n = len(nums)
        p = [0] * (n+1)

        for s,e in requests:
            p[s] += 1
            p[e+1] -= 1

        for i in range(1, n+1):
            p[i] += p[i-1]

        nums.sort(reverse = True)
        p.sort(reverse = True)
        res = 0
        for i,x in enumerate(nums):
            res += x * p[i]
            res %= M

        return res